Find the value of p for which the quadratic equation x^2 - px + p+ 3=0 has a) coincident roots b) real distinct roots c) one positive and one negative root. Please solve ☺😊😇😄

Dear Student,
Please find below the solution to the asked query:

We have,x2-px+p+3=0a For coincident roots or equal roots,D=0-p2-4p+3=0p2-4p-12=0p2-6p+2p-12=0pp-6+2p-6=0p-6p+2=0p=6 or p=-2b For real distinct roots,D>0-p2-4p+3>0p2-4p-12>0p2-6p+2p-12>0pp-6+2p-6>0p-6p+2>0p>6 or p<-2p-,-26,c For one negative and one positive rootD>0 and 1×f0<0-p2-4p+3>0 and p+3<0p>6 or p<-2 and p<-3      From bp<-3

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